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I am interested in a closed form expression for the convex conjugate $$ f^*(y) = \sup_{x \ge 0} x y - f(x) $$ of the function $$ f \colon \mathbb R \to \mathbb R, \qquad x \mapsto \begin{cases} |1 - x^{a} |^{\frac{1}{a}}, & \text{if } x \ge 0\\ \infty, & \text{else.} \end{cases}, $$ where $a \in (0, 1)$.

I think that this is not possible because in the intersection of interior of the domain of $f$ and the set of its differentiability points, we should have the following: to find the maximiser in the expression of $f^*$ we take the derivative and set it to zero, resulting in $y - f'(x) = 0$, that is $x = (f')^{-1}(y)$.

We have $$f'(y) = y^{a - 1} (y^a - 1) | 1 - y^a |^{\frac{1}{a} - 2} \qquad \forall y \in (0, 1) \cup (1, \infty).$$ However, there does not seem be an explicit / closed for for $(f')^{-1}$.

Is my reasoning correct?

ViktorStein
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1 Answers1

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Notice we can use the sign function and rearrange the powers: for $1\ne y>0$ and $0<a<1$ we have \begin{align*} g & =f'(y) = y^{a - 1} (y^a - 1) |y^a-1|^{\frac{1}{a} - 2} =(y^{-a})^{\frac1a-1}\frac{y^a-1}{|y^a-1|}|y^a-1|^{\frac1a-1} \\ & =\text{sgn}(y^a-1)\left|\frac1{y^a}-1\right|^{\frac1a-1}. \end{align*} Further, $\text{sgn}(y^a-1)=\text{sgn}(y-1)$ can be treated like a constant as its range is $\{\pm1\}$, so now we solve for $g$: $$y=\text{sgn}(g-1) \left|\frac1{g^a}-1\right|^{\frac1a-1}\iff (\mp y)^\frac a{1-a}=\left|\frac1{g^a}-1\right|\\\pm(\mp y)^\frac a{1-a}=\frac1{g^a}-1$$

$(-y)^\frac a{1-a}>0,y<0$ and $y^\frac a{1-a}>0,y>0$. Also, graphically, there are alternating signs. Technically, $y\ne0$ as $(f’)^{-1}(0)=1$ and $f’(1)$ was not defined. Therefore:

$$\boxed{f(x)=|1-x^a|^\frac1a\implies (f’)^{-1}(y)=\begin{cases}\left(1+(-y)^\frac a{1-a}\right)^{-\frac1a}&y<0\\ \left(1-y^\frac a{1-a}\right)^{-\frac1a}&0<y<1.\end{cases}}$$

shown here. For certain $a$, one of either branches works on all of $0\ne y<1$.

Тyma Gaidash
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